What is the speed of the block in this compressed spring problem?

In summary, the problem presented is about a block and spring on a frictionless incline. The spring is initially compressed by 0.35 m and then released. The question asks for the speed of the block when the spring is compressed by 0.14 m. The solution involves conservation of mechanical energy and calculating the gravitational potential energy using the distance d = 0.14 m and angle 30 degrees. The final velocity is found to be 4.79 m/s.
  • #1
BlackMamba
187
0
I have a problem that I just can't seem to figure out.

Here's the problem: The drawing (Click Here) shows a block (m = 1.6 kg) and a spring (k = 325 N/m) on a frictionless incline. The spring is compressed by x_o = 0.35 m relative to it's unstrained position at x = 0 m and then released. What is the speed of the block when the spring is still compressed by x_f = 0.14 m?

I applied the whole Ef = Eo equation but did not get the right answer.

I used sin 30 to find h_f.

So this is what I used:
V_i = 0
V_f = what I'm looking for
h_i = 0
h_f = .5
k = 325
x_i = 0.35
x_f = 0.14


Any help on this problem would be greatly appreciated.

Thanks in advance.
 
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  • #2
This is a problem of application of conservation of mechanical energy

[tex] \Delta \Omega + \Delta K = 0 [/tex]

The energy involved here are kinetic energy, spring potential energy and gravitational potential energy.

At the beginning the spring has spring potential energy and gravitational potential energy, when it reaches the point on the diagram it has spring potential energy and kinetic energy.
 
  • #3
Alrighty.

I know that:
KE = 1/2 mv^2
Spring PE = 1/2 kx
Gravitational PE = mgh

So my equation would look like: 1/2 kx_i + mgh = 1/2 kx_f + 1/2 mv^2

Am I calculating h correctly by using sin 30?

Thanks
 
  • #4
If you take the 0.14 as your referend for the gravitational potential enegry you should have:

[tex] info [/tex]
[tex] x=0.35 [/tex]
[tex] d=0.14 [/tex]
[tex] \theta = 30^{o} [/tex]

[tex] \frac{1}{2}kx^2 + mg(x-d) \sin \theta = \frac{1}{2}kd^2 + \frac{1}{2}mv^2 [/tex]
 
Last edited:
  • #5
Ok well at least I was close. LOL but I don't understand what d is?
 
  • #6
Well, I tried the equation that you gave with those values you specified and I still am not getting the right answer. There must be something I am missing.

The answer I got was 2.33 m/s and it is wrong.

Thanks for helping.
 
  • #7
I am not getting your numeric value, i get 4.79 m/s
 

1. What is the "Compressed Spring Problem"?

The "Compressed Spring Problem" is a physics problem that involves calculating the displacement of a spring when a force is applied to it. It is often used to demonstrate the relationship between force, displacement, and spring constants.

2. How do you solve the "Compressed Spring Problem"?

To solve the "Compressed Spring Problem", you can use the equation F = kx, where F is the applied force, k is the spring constant, and x is the displacement of the spring. You can also use the equation U = 1/2kx^2, where U is the potential energy of the spring.

3. What are some real-life applications of the "Compressed Spring Problem"?

The "Compressed Spring Problem" has many real-life applications, such as in shock absorbers, car suspensions, and pogo sticks. It is also used in the design of springs for various mechanical and electrical devices.

4. What factors affect the displacement of a compressed spring?

The displacement of a compressed spring is affected by the applied force, the spring constant, and the initial length of the spring. Other factors, such as the material and thickness of the spring, can also affect its displacement.

5. How does the "Compressed Spring Problem" relate to Hooke's Law?

The "Compressed Spring Problem" is directly related to Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring. The "Compressed Spring Problem" is a practical application of Hooke's Law.

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